Câu hỏi của Đặng Hà Minh Huyền

Question

Cho x=$\sqrt{4-\sqrt{10+2\sqrt{5}}}$ và y=$\sqrt{4+\sqrt{10+2\sqrt{5}}}$

Tính x.y; x+y

Lê Bùi · Accepted Answer

X+Y=$\sqrt{4-\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}$

$\left(X+Y\right)^2=(\sqrt{4-\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10+2\sqrt{5}}})^2$

$\left(X+Y\right)^2=4-\sqrt{10+2\sqrt{5}}+4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4-\sqrt{10+2\sqrt{5}}\right)\left(4+\sqrt{10+2\sqrt{5}}\right)}$

$\left(X+Y\right)^2=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}$

$\left(X+Y\right)^2=8+2\sqrt{6-2\sqrt{5}}$

$\left(X+Y\right)^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}$

$\left(X+Y\right)^2=8+2\sqrt{5}-2$

$\left(X+Y\right)^2=6+2\sqrt{5}$

$X+Y=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1$Câu trả lời: X+Y=$\sqrt{4-\sqrt{10+2\sqrt{5}}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}$